A jet airliner moving initially at 521 mph
(with respect to the ground) to the east moves
into a region where the wind is blowing at
434 mph in a direction 22◦
north of east.
What is the new speed of the aircraft with
respect to the ground?
To find the new speed of the aircraft with respect to the ground, we can use vector addition.
1. Start by representing the velocity of the aircraft as a vector. Since it is moving initially at 521 mph to the east, the vector representing its velocity will have a magnitude of 521 mph in the east direction.
2. Represent the velocity of the wind as another vector. The wind is blowing at 434 mph in a direction 22° north of east. To represent this vector, we can break it into its east and north components. The east component can be found using the cosine function: 434 mph * cos(22°). The north component can be found using the sine function: 434 mph * sin(22°).
3. Add the two vectors together. Add the east component of the wind velocity vector to the east component of the aircraft velocity vector, and add the north component of the wind velocity vector to the north component of the aircraft velocity vector. This will give you the new velocity vector of the aircraft with respect to the ground.
4. Find the magnitude of the new velocity vector. This will give you the new speed of the aircraft with respect to the ground.
Let's do the calculations:
East component of wind velocity = 434 mph * cos(22°) = 408.75 mph
North component of wind velocity = 434 mph * sin(22°) = 159.40 mph
New east velocity component = 521 mph + 408.75 mph = 929.75 mph
New north velocity component = 159.40 mph
So, the new velocity of the aircraft with respect to the ground is a vector with an east component of 929.75 mph and a north component of 159.40 mph.
To find the magnitude of this vector, we can use the Pythagorean theorem:
New speed = sqrt((929.75 mph)^2 + (159.40 mph)^2) = 947.21 mph
Therefore, the new speed of the aircraft with respect to the ground is approximately 947.21 mph.